Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
const int N = 1e5 + 10;
vector<int> order;
const int LOG = 18;
int L[N];
int p[N][LOG];
int par[N];
vector<int> g[N];
vector<int> lives[N][LOG];
vector<int> pessoas[N];
vector<int> merge(vector<int> &a, vector<int> &b, int k) {
  int i, j;
  i = j = 0;
  vector<int> ans;
  for (int i = 0, j = 0; i < a.size() || j < b.size();) {
    if (i == a.size()) {
      ans.push_back(b[j++]);
    } else if (j == b.size()) {
      ans.push_back(a[i++]);
    } else if (a[i] == b[j]) {
      ans.push_back(a[i++]);
      j++;
    } else if (a[i] < b[j]) {
      ans.push_back(a[i++]);
    } else if (b[j] < a[i]) {
      ans.push_back(b[j++]);
    }
    if (ans.size() == k) break;
  }
  return ans;
}
void dfs(int x, int dad) {
  p[x][0] = dad;
  for (int i = 1; i < LOG; i++) {
    p[x][i] = p[p[x][i - 1]][i - 1];
    lives[x][i] = merge(lives[x][i - 1], lives[p[x][i - 1]][i - 1], 10);
  }
  for (auto u : g[x]) {
    if (u == dad) continue;
    lives[u][0] = pessoas[x];
    L[u] = L[x] - 1;
    dfs(u, x);
  }
}
int lca(int x, int y, vector<int> &ans, int k) {
  if (L[x] > L[y]) {
    swap(x, y);
  }
  ans = merge(pessoas[x], pessoas[y], k);
  for (int i = LOG - 1; i >= 0; i--) {
    if (L[x] + (1 << i) <= L[y]) {
      ans = merge(ans, lives[x][i], k);
      x = p[x][i];
    }
  }
  if (x == y) {
    return x;
  }
  for (int i = LOG - 1; i >= 0; i--) {
    if (p[x][i] != p[y][i]) {
      ans = merge(lives[x][i], ans, k);
      ans = merge(lives[y][i], ans, k);
      x = p[x][i];
      y = p[y][i];
    }
  }
  return p[x][0];
}
vector<int> solve(int a, int b, int k) {
  vector<int> ans;
  int LCA = lca(a, b, ans, k);
  return merge(pessoas[LCA], ans, k);
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    scanf("%d", &a);
    a--;
    if (pessoas[a].size() < 10) {
      pessoas[a].push_back(i);
    }
  }
  dfs(0, 0);
  while (q--) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--;
    b--;
    vector<int> ans = solve(a, b, c);
    printf("%d", ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf(" %d", ans[i]);
    }
    puts("");
  }
  return 0;
}